the sum of the square of two consecutive positive integers is 265.what are the integers?
Since "square" is singular, I'm not sure which one you mean:
(x + x + 1)^2 = 265
x^2 + (x+1)^2 = 265
Either way, solve for x.
To solve this problem, let's assume the two consecutive positive integers are x and x + 1.
According to the problem, the sum of the square of these two consecutive positive integers is 265. So we can write the equation as:
x^2 + (x + 1)^2 = 265
Now let's solve this equation and find the values of x and x + 1.
Expanding the equation:
x^2 + (x^2 + 2x + 1) = 265
Combining like terms:
2x^2 + 2x + 1 = 265
Subtracting 265 from both sides:
2x^2 + 2x + 1 - 265 = 0
Simplifying:
2x^2 + 2x - 264 = 0
Dividing the equation by 2 to simplify it further:
x^2 + x - 132 = 0
Now let's factorize the equation:
(x + 12)(x - 11) = 0
Setting each factor equal to zero and solving for x:
x + 12 = 0 or x - 11 = 0
If x + 12 = 0, then x = -12. But since we are looking for positive integers, this is not a valid solution.
If x - 11 = 0, then x = 11. This is a valid solution.
So, the two consecutive positive integers are 11 and 12.